# A question about direct product

Let $G$ be a group with subgroups $G_1,..., G_n$ such that $\mathcal{D}_1,\mathcal{D}_2,\mathcal{D}_3$ hold, then the mapping \begin{align*} \alpha: {\huge\times}_{i=1}^n G_i\rightarrow G\text{ with }(g_1,...,g_n)\mapsto g_1\cdots g_n \end{align*} is a isomorphism.

$\mathcal{D}_1~~~~G=G_1\cdots G_n;$

$\mathcal{D}_2~~~~G_i\trianglelefteq G;$

$\mathcal{D}_3~~~~G_i\cap \prod_{j\neq i}G_j=1.$

It's a rather common proposition. But I'm stuck on one point when I tried to go through the proof.

Question:

$G_i\cap \prod_{j\neq i}G_j$ yields $[G_i,G_j]\leq G_i\cap\prod_{j\neq i}G_j=1$, which means for all $1\leq i,j\leq n$, $G_i,G_j$ centralizes each other. Since all $g\in G$ can be written as $g=g_1\cdots g_n$, then we'll have $$\boldsymbol{ gg^*=(g_1\cdots g_n)(g^*_1\cdots g^*_n)=(g^*_1\cdots g^*_n)(g_1\cdots g_n)=g^*g,}$$ where $g,g^*$ are arbitrary elements in $G$.

But it is utterly ridiculous, isn't it?

I'm a beginner in group theory, would you please give me some help and point out where I misunderstood? Thanks!

• You seem to be assuming that $g_i$ commutes with $g_i^*$. That's not necessarily true if $G_i$ is nonabelian. – Bungo Mar 13 '18 at 5:38
• @Bungo Ah, yes! I feel like a fool... Thanks a lot. – Math Mar 13 '18 at 5:40
• Should I delete this post? It this question valuable? – Math Mar 13 '18 at 5:41
• No problem. The rest of your argument, namely that elements of $G_i$ commute with elements of $G_j$ whenever $i \neq j$, is correct. – Bungo Mar 13 '18 at 5:41
• Someone might find it useful. I'll go ahead and write a quick answer so it doesn't have to stay open. – Bungo Mar 13 '18 at 5:43

Your argument is almost correct: elements of $G_i$ commute with elements of $G_j$ provided that $i \neq j$. However, elements of $G_i$ need not commute with each other, since $G_i$ is not assumed abelian.
• So the positions of $G_i$ can change, am I right? – Math Mar 13 '18 at 5:56
• If you mean that we can reorder the $G_i$'s in the product $G = G_1 G_2 \cdots G_n$, that is correct. In fact, you only need hypothesis $\mathcal D_2$ in order to do this: normal subgroups $H$ and $K$ always satisfy $HK = KH$, and by induction this holds for any finite number of normal subgroups. Your result is stronger: not only do the subgroups $G_i$ commute as sets, their individual elements also commute. – Bungo Mar 13 '18 at 5:59